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01 Feb 2021, 03:00
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25% (02:45) correct
75%
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If \(||x - 1| + |x|| = |x + 3| - x\), then what is the value of x ?
(1) \(|x|\) is not a prime number
(2) \(2x^2\) is a prime number
(1) \(|x|\) is not a prime number
(2) \(2x^2\) is a prime number
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01 Feb 2021, 08:56
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Bunuel
\(||x - 1| + |x|| = |x + 3| - x\)
The critical points are -3, 0 and 1, so three cases
1) \(x<-3\)
\(||x - 1| + |x|| = |x + 3| - x.........|1 -x-x|=-x-3-x.......|1-2x|=-3-2x........1-2x=-3-2x....-1=-3\)...Not possible
2) \(0< x\geq{-3}\)
\(||x - 1| + |x|| = |x + 3| - x.........|1 -x-x|=x+3-x.......|1-2x|=3........1-2x=3........2x=-2....x=-1\)
3) \(1< x\geq{0}\)
\(||x - 1| + |x|| = |x + 3| - x.........|1 -x+x|=x+3-x.......1=3........\) Not possible
4) \( x\geq{1}\)
\(||x - 1| + |x|| = |x + 3| - x.........|x-1+x|=x+3-x.......|2x-1|=3........2x-1=3........2x=4.......x=2\)
So two possible values of x are -1 and 2.
(1) \(|x|\) is not a prime number
Possible value is -1
Sufficient
(2) \(4x^2\) is a prime number
Something is missing in the choice. In the given scenario, 4x^2 will never be a prime number, unless we have x as \(\frac{\sqrt{prime}}{2}\) and here we have a linear equation so square root is not possible.
Instead of 4x^2, it should be 2x^2 or 3x^2 or 5x^2.
In that case only -1 is possible
Sufficient
General Discussion
Poojita
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06 Apr 2021, 08:01
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chetan2u
why havent we tried out values from the ranges i.e -infinity to -3 then from -3 to 0, 0 to 1 and 1 to infinity?
why did you approached it using variables?
Could you also explain how statement B is sufficient?
